# Integral of $\int_0^{\pi/2} \ (\sin x)^7\ (\cos x)^5 \mathrm{d} x$

I am trying to find this by using integration by parts but I am not sure how to do it.

$$\int_0^{\pi/2} (\sin x)^7 (\cos x)^5 \mathrm{d} x$$

I tried rewriting as

$$\int_0^{\pi/2} \sin x\cdot\ (\sin x)^6\cdot\ (\cos x)^5 \mathrm{d} x = \int_0^{\pi/2}\sin x(1-\ (\cos x)^3)\cdot\ (\cos x)^5 \mathrm{d} x$$

but that seems to only give me a very, very long loop that doesn’t help me at all. How do I proceed?

$$\int_0^{\pi/2} \sin x\cdot (\sin x)^6\cdot (\cos x)^5 \mathrm{d} x = \sin x(1- (\cos x)^3)\cdot (\cos x)^5 \mathrm{d} x$$

$u = \cos x$, then $du = -\sin xdx$

$\int \frac{-u^6}{6} \mathrm{d} u – \int \frac{-u^9}{9} \mathrm{d} u$

From here it looks like I have an incredibly long string of $u$ substitutions to make to get to something I can find an antiderivative for.

#### Solutions Collecting From Web of "Integral of $\int_0^{\pi/2} \ (\sin x)^7\ (\cos x)^5 \mathrm{d} x$"

To integrate the function $$f(x)=\sin ^{n}x\cdot\cos ^{m}x,$$ when $n$ or $m$ are positive odd numbers, we can apply a general technique which consists of expanding $f(x)$ into a sum of terms of the form $$\sin ^{p}x\cdot \cos x,\qquad p=1,2,\ldots$$ or $$\cos ^{q}x\cdot \sin x,\qquad q=1,2,\ldots.$$ Using the
identity
$$\begin{equation*} \cos ^{2}x=1-\sin ^{2}x, \end{equation*}$$
in the form
$$\begin{equation*} \cos ^{4}x=(1-\sin ^{2}x)^2 \end{equation*}=1-2\sin ^{2}x+\sin ^{4}x,$$

we rewrite our $$f(x)=\sin ^{7}x\cdot\cos ^{5}x=\sin ^{7}x\cdot\cos ^{4}x\cdot\cos x$$ as
$$\begin{eqnarray*} f(x) &=&\sin ^{7}x\cdot \left( 1-2\sin ^{2}x+\sin ^{4}x\right) \cdot \cos x \\ &=&\sin ^{7}x\cdot \cos x-2\sin ^{9}x\cdot \cos x+\sin ^{11}x\cdot \cos x. \end{eqnarray*}$$

Each term is of the form $\sin ^{p}x\cdot \cos x$ and can easily be
integrated by the substitution $u=\sin x$, $u^{\prime }=\cos x$, $du=\cos x\;dx=u’\;dx$:
$$\begin{eqnarray*} \int \sin ^{p}x\cdot \cos x\;dx &=&\int u^{p}\;du=\frac{u^{p+1}}{p+1}=\frac{\sin ^{p+1}x}{p+1}+C, \\ \int_{0}^{\pi /2}\sin ^{p}x\cdot \cos x\;dx &=&\frac{1}{p+1}. \end{eqnarray*}$$

Added: detailed computation in view of OP’s comment

$$\begin{eqnarray*} \int_{0}^{\pi /2}f(x)dx &=&\int_{0}^{\pi /2}\sin ^{7}x\cos ^{5}xdx \\ &=&\int_{0}^{\pi /2}\sin ^{7}x\cdot \cos xdx-2\int_{0}^{\pi /2}\sin ^{9}x\cdot \cos xdx \\ &&+\int_{0}^{\pi /2}\sin ^{11}x\cdot \cos xdx \\ &=&\frac{1}{8}-2\cdot \frac{1}{10}+\frac{1}{12}=\frac{1}{120}. \end{eqnarray*}$$

If …you actually meant $\,\,\displaystyle{\int_0^{\pi/2}\sin^7x\cos^5x\,\, dx}\,\,$ then, putting $\,s:= \sin x\,\,,\,c:=\cos x\,$ , we get:$$\int s^7c^5=\frac{1}{2}\int 2sc(1-s^2)^3c^4=\frac{1}{8}c^4(1-c^2)^4+\frac{1}{2}\int c^3s(1-c^2)^4 –\text{ int. by parts, with}$$$$\,u=c^4\,,\,u’=-4c^3s\,\,,\,\,v’=2sc(1-c^2)^3\,,\,v=\frac{(1-c^2)^4}{4}$$ Here we can again put things as before within the integral: $$c^3s(1-c^2)^4=\frac{1}{2}\left[2cs(1-c^2)^4c^2\right]$$and do integration by parts again. I’ll leave this for you.

Integration by parts is not the method to use here. $u$-substitution is the method to use. This is perhaps not so obvious, except that as you noticed, IBP doesn’t seem to get you very far right off.

You have $\displaystyle \int \sin x (1 – \cos^3 x) \cos ^5 x dx = \int \sin x \cos ^5 x dx – \int \sin x \cos ^8 x dx$

For the first, note that $-\sin x$ is the dervative of $\cos x$, and use $u$-substitution. Do the same for the second. Does that make sense?

EDIT

I see now that the rewritten formula in the OP is not actually correct. $\sin^6 x = (\sin^2 x)^3 = (1 – \cos ^2 x)^3$. So you must substitute this (instead of $1 – \cos ^3$) into the integral and proceed. My work above still gives the way to the answer. You might also work from $\sin x \sin^4 x$ instead – and it might even save you some time.

Further Edited for an example

I see some confusion. But let’s suppose we had $\int \sin x \cos^2 x dx$. I know that $-\sin x = \frac{d}{dx} \cos x$, so if I let $u = \cos x$, $du = -\sin x dx$, then we have that

$$\int \sin x \cos ^2 x dx = -\int u^2 du$$

And we know how to calculate this.

$$-\int u^2 du = -\frac{u^3}{3} + C$$

As $u = \cos x$, ths actually says that

$$\int \sin x \cos ^2 x dx = -\frac{\cos^3 x}{3} + C$$

So in this way, there is not a string of $u$-substitutions, but just one. Your integrals can be handled similarly. I encourage you to update us with your work.

This type of integrals can be solved to give the general result:

$$\int_0^{\pi /2}\sin^m\theta\cos^n\theta d \theta=\begin{cases} \frac{(m-1)!!(n-1)!!}{(m+n)!!} \text{ if any exponent is odd}\cr \frac{(m-1)!!(n-1)!!}{(m+n)!!}\frac{\pi} 2 \text{ both even exponents} \end{cases}$$

We first prove the reduction formula:

$$\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}xdx} = \frac{{m – 1}}{{m + n}}\int\limits_0^{\pi /2} {{{\sin }^{m – 2}}x{{\cos }^n}xdx}$$

This is done by integrating by parts with $\sin^{m-1} x=v$ and $\cos^n x \sin x dx = du$, which gives

\eqalign{ & \int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}xdx} = \frac{{m – 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m – 2}}x{\cos ^2}xdx \cr & \int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}xdx} = \frac{{m – 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m – 2}}x\left( {1 – {{\sin }^2}x} \right)dx \cr & \int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}xdx} = \frac{{m – 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m – 2}}xdx – \frac{{m – 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^m}xdx \cr & \left( {1 + \frac{{m – 1}}{{n + 1}}} \right)\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}x\sin xdx} = \frac{{m – 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m – 2}}xdx \cr & \frac{{m + n}}{{n + 1}}\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}x\sin xdx} = \frac{{m – 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m – 2}}xdx \cr & \int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}x\sin xdx} = \frac{{m – 1}}{{m + n}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m – 2}}xdx \cr}

With this proved, we want to get to an easier integral. The pattern is evident:

$$\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}x\sin xdx} = \frac{{m – 1}}{{m + n}}\frac{{m – 3}}{{m + n – 2}} \cdots \frac{{m – 2k + 1}}{{m + n – 2k + 2}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m – 2k}}xdx$$

So what we want now is $2k=m$. We get

$$\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}x\sin xdx} = \frac{{\left( {m – 1} \right)!!}}{{m + n}}\frac{1}{{m + n – 2}} \cdots \frac{1}{{n + 2}}\int\limits_0^{\pi /2} {{{\cos }^n}x} dx$$

so it all burns down to finding

$$\int\limits_0^{\pi /2} {{{\cos }^n}x} dx$$

In the same spirit as before, we integrate by parts, reducing the power of the cosine:

\eqalign{ & \int\limits_0^{\pi /2} {{{\cos }^n}x} dx = \left( {n – 1} \right)\int\limits_0^{\pi /2} {{{\cos }^{n – 2}}x{{\sin }^2}xdx} \cr & \int\limits_0^{\pi /2} {{{\cos }^n}x} dx = \left( {n – 1} \right)\int\limits_0^{\pi /2} {{{\cos }^{n – 2}}x\left( {1 – {{\cos }^2}x} \right)dx} \cr & \int\limits_0^{\pi /2} {{{\cos }^n}x} dx = \left( {n – 1} \right)\int\limits_0^{\pi /2} {{{\cos }^{n – 2}}xdx} – \left( {n – 1} \right)\int\limits_0^{\pi /2} {{{\cos }^n}xdx} \cr & n\int\limits_0^{\pi /2} {{{\cos }^n}x} dx = \left( {n – 1} \right)\int\limits_0^{\pi /2} {{{\cos }^{n – 2}}xdx} \cr & \int\limits_0^{\pi /2} {{{\cos }^n}x} dx = \frac{{n – 1}}{n}\int\limits_0^{\pi /2} {{{\cos }^{n – 2}}xdx} \cr}

Depending on wether $n$ is or isn’t even, we’ll end up with

$$\int\limits_0^{\pi /2} {{{\cos}^n}x} dx =\begin{cases} \frac{{\left( {n – 1} \right)!!}}{{n!!}} \frac{\pi} 2 \text{ n even} \cr \frac{{\left( {n – 1} \right)!!}}{{n!!}} \text{ n odd} \end{cases}$$

Since the last factor will be either $${\int\limits_0^{\pi /2} {dx} }$$ or $${\int\limits_0^{\pi /2} {\cos xdx} }$$

You can easily show the same symmetric results, i.e.

$$\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}xdx} = \frac{{n – 1}}{{m + n}}\int\limits_0^{\pi /2} {{{\sin }^{m }}x{{\cos }^{n-2}}xdx}$$

and

$$\int\limits_0^{\pi /2} {{{\sin}^n}x} dx =\begin{cases} \frac{{\left( {n – 1} \right)!!}}{{n!!}} \frac{\pi} 2 \text{ n even} \cr \frac{{\left( {n – 1} \right)!!}}{{n!!}} \text{ n odd} \end{cases}$$

“Gluing” all this together, we get the first stated result

$$\int_0^{\pi /2}\sin^m\theta\cos^n\theta d \theta=\begin{cases} \frac{(m-1)!!(n-1)!!}{(m+n)!!} \text{ if any exponent is odd}\cr \frac{(m-1)!!(n-1)!!}{(m+n)!!}\frac{\pi} 2 \text{ both even exponents} \end{cases}$$

so

$$\int\limits_0^{\pi /2} {{{\sin }^7}x{{\cos }^5}x\sin xdx} = \frac{{\left( {7 – 1} \right)!!\left( {5 – 1} \right)!!}}{{\left( {7 + 5} \right)!!}} = \frac{1 }{{120}}$$

ADD: By letting $m=2y-1$ and $n=2x-1$, we get the famous Beta integral for integer values:

$$\int_0^{\pi /2} {{{\sin }^{2y – 1}}} \theta {\cos ^{2x – 1}}\theta d\theta = \frac{{(2y – 2)!!(2x – 2)!!}}{{(2x – 1 + 2y – 1)!!}}{\text{ = }}\frac{{{2^{y – 1}}\left( {y – 1} \right)!{2^{x – 1}}(x – 1)!}}{{{2^{y + x – 1}}(x + y – 1)!}}{\text{ = }}\frac{1}{2}\frac{{\left( {y – 1} \right)!(x – 1)!}}{{(x + y – 1)!}}$$

Beta version:

\begin{aligned}\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{7}\left(\cos x\right)^{5}dx & =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{6}\left(\cos x\right)^{4}2\sin x\cos xdx\\ & =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\left(\sin x\right)^{2}\right)^{3}\left(1-\left(\sin x\right)^{2}\right)^{2}d\left(\left(\sin x\right)^{2}\right)\\ & =\frac{1}{2}\int_{0}^{1}t^{3}\left(1-t\right)^{2}dt=\frac{1}{2}\int_{0}^{1}t^{4-1}\left(1-t\right)^{3-1}dt\\ & =\frac{1}{2}B\left(4,3\right)=\frac{1}{2}\frac{\Gamma\left(4\right)\Gamma\left(3\right)}{\Gamma\left(7\right)}=\frac{1}{2}\frac{6\cdot2}{720}=\frac{1}{120} \end{aligned}

Let I $= \int_0^{\pi/2} \ (\sin x)^7\ (\cos x)^5 \mathrm{d} x$

Then using the substitution u=${\pi/2}-x$ we have I $= \int_0^{\pi/2} \ (\sin x)^5\ (\cos x)^7 \mathrm{d} x$

Adding gives $$2I= \int_0^{\pi/2} \ (\sin x)^5\ (\cos x)^5 \mathrm{d} x$$

Since $cos^2(x)+sin^2(x) = 1$

And that $$2I= {1\over2^5}\ \int_0^{\pi/2}(\sin 2x)^5 \mathrm{d} x$$

since $sin(2x)=2sin(x)cos(x)$. Then substituting u=2x gives

$$2I= {1\over2^6}\ \int_0^{\pi}(\sin u)^5 \mathrm{d} u = {1\over2^6}\ \int_0^{\pi}(\sin u) (1-\cos^2 u)^2 \mathrm{d} u$$

$$= {1\over2^6}\ \int_0^{\pi}(\sin u) (1-2\cos^2 u+\cos^4(u)) \mathrm{d} u$$

$$= {1\over2^6}\ \int_0^{\pi}(\sin u) (1-2\cos^2 u+\cos^4(u)) \mathrm{d} u$$

Hence $$I = {1\over2^7}\ [{-\cos u}+ {2\cos^3 u\over 3}- {\cos^5 u\over 5}]_0^\pi$$

$$I = {1\over2^7}\ [{2}+ {-2\ 2 \over 3}+ {2 \ \over 5}]_0^\pi$$

Hence $$I = {1\over{2^3}\ 15}={1\over120}= {1\over{2^3}\ 15}={1\over120}$$